Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $p = \dfrac{-2q^2 + 4q + 160}{q^2 - 11q + 10} \div \dfrac{q^2 + 5q}{q^2 - q} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-2q^2 + 4q + 160}{q^2 - 11q + 10} \times \dfrac{q^2 - q}{q^2 + 5q} $ First factor out any common factors. $p = \dfrac{-2(q^2 - 2q - 80)}{q^2 - 11q + 10} \times \dfrac{q(q - 1)}{q(q + 5)} $ Then factor the quadratic expressions. $p = \dfrac {-2(q - 10)(q + 8)} {(q - 10)(q - 1)} \times \dfrac {q(q - 1)} {q(q + 5)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac { -2(q - 10)(q + 8) \times q(q - 1)} { (q - 10)(q - 1) \times q(q + 5)} $ $p = \dfrac {-2q(q - 10)(q + 8)(q - 1)} {q(q - 10)(q - 1)(q + 5)} $ Notice that $(q - 10)$ and $(q - 1)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {-2q\cancel{(q - 10)}(q + 8)(q - 1)} {q\cancel{(q - 10)}(q - 1)(q + 5)} $ We are dividing by $q - 10$ , so $q - 10 \neq 0$ Therefore, $q \neq 10$ $p = \dfrac {-2q\cancel{(q - 10)}(q + 8)\cancel{(q - 1)}} {q\cancel{(q - 10)}\cancel{(q - 1)}(q + 5)} $ We are dividing by $q - 1$ , so $q - 1 \neq 0$ Therefore, $q \neq 1$ $p = \dfrac {-2q(q + 8)} {q(q + 5)} $ $ p = \dfrac{-2(q + 8)}{q + 5}; q \neq 10; q \neq 1 $